Read a number, show me a word!

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Hello!!!
Well, I want to make a programme in C which will do this:

I will type 1 and it will appear "one" on the screen.
If I had a small number of variables I could do it like this:

C#

switch (x)

{
  case 0:printf("Zero.\n");
         break;
  case 1:printf("One.\n"); etc...
         break;
}

But in my programme i wanna do this with the numbers 0 to 99!!
I don't think that I should write a hundred cases after the switch!(at list I hope so...)

Thank you in advance!! :-)

Posted 14-Nov-11 8:57am

MG2011

Updated 14-Nov-11 9:00am

v2

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Sergey Alexandrovich Kryukov 14-Nov-11 14:59pm

What did you try? Forget your switch... any serious ideas? This is so simple.
--SA

MG2011 14-Nov-11 15:02pm

I did't think of anything else... :-(

4 solutions

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Andrew Brock · Answer 1 · 2011-11-14T18:47:00

This is quite simple.
If all you want is [0-99] then all you need is the GetHundredsWords below, but this code will print any number you can put in an integer.

For every order of 1000, the numbering is the same, for example: 123 is one hundred twenty three and 123,000 is one hundred twenty three thousand, so we only really need code which can go up to 1000 (GetHundredsWords()), then just do that for every order of 1000, adding in the correct magnitude (GetWord())

C++

#include <stdio.h>

static size_t GetHundredsWords(char *buffer, int n, bool &bAppended) {
	//Shorthand to make sure spaces are added when needed
	#define APPEND(fmt, ...) \
		if (bAppended) { \
			*ptr++ = ' '; \
		} else { \
			bAppended = true; \
		} \
		ptr += sprintf(ptr, fmt, __VA_ARGS__);

	const char *szOnesWords[] = { "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };
	const char *szTeenWords[] = { "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };
	const char *szTensWords[] = { "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" };
	char *ptr = buffer;
	if (n >= 100) {
		APPEND("%s hundred", szOnesWords[n / 100 - 1]);
		n %= 100;
	}
	if (n >= 20) {
		APPEND(szTensWords[n / 10 - 2]);
		n %= 10;
	}
	if (n >= 10) {
		APPEND(szTeenWords[n - 10]);
	} else if (n > 0) {
		APPEND(szOnesWords[n - 1]);
	}
	return ptr - buffer;
}

char *GetWord(char *buffer, int n) {
	const char *szMagnitudes[] = { " billion", " million", " thousand", "" };
	char *ptr = buffer;
	if (n == 0) {
		return "zero"; //Special case
	}
	if (n < 0) { //Is this a negative number
		ptr += sprintf(ptr, "negative ");
		n *= -1;
	}
	bool bAppended = false;
	int nMagnitude = 1000 * 1000 * 1000; //1 billion
	for (int i = 0; nMagnitude > 0; ++i) {
		if (n >= nMagnitude) {
			ptr += GetHundredsWords(ptr, n / nMagnitude, bAppended);
			ptr += sprintf(ptr, szMagnitudes[i]);
			n %= nMagnitude;
		}
		nMagnitude /= 1000;
	}
	//GetHundredsWords(ptr, n, bAppended, bAnd);
	return buffer;
}

#define TEST(n) printf("%d is \'%s\'\n", n, GetWord(buffer, n)) //Save having to type this for every test

int main(int argc, char *argv[]) {
	char buffer[64];
	TEST(0);
	TEST(1);
	TEST(5);
	TEST(9);
	TEST(10);
	TEST(11);
	TEST(12);
	TEST(19);
	TEST(20);
	TEST(31);
	TEST(42);
	TEST(100);
	TEST(101);
	TEST(109);
	TEST(110);
	TEST(115);
	TEST(120);
	TEST(151);
	TEST(546);
	TEST(1000);
	TEST(1001);
	TEST(1024);
	TEST(1564);
	TEST(1000000);
	TEST(1000000000);
	TEST(1561981328);
	TEST(-2147483647);
	return 0;
}

lewax00 · Answer 2 · 2011-11-14T10:33:00

You'll probably want to break it down into ranges, for example anything between 20 and 29 will start with "twenty", so something like:

C++

if (number >= 20 && number <= 29)
{
    printf("Twenty");
}

Of course, you'd need something similar for 30, 40, and so on

Then you'll want to print something after that depending on the last digit, there are a few different ways to do that, like the modulus operator, or if you're breaking down into ranges like this you can use subtraction (e.g. if its in the range of 20's, subtract twenty and all you'll have left if the last digit, 21 goes to 1, 22 goes to 2, etc.) So after printing the first word (if there is one) you can then print the remaining word (using a switch for this last part could work).

Don't forget special cases, for example if it's just 0 you'd want it to print "Zero" but if it's 20 you wouldn't want "Twenty Zero"!

Sergey Alexandrovich Kryukov · Answer 3 · 2011-11-14T09:04:00

Solution 1

Never ever use a switch which contains just a few cases. And in your sample, even two cases would be a total abuse.

You need just one function which accepts an integer in certain range and returns a string. If a value is outside the allowed range, return, for example, null. Do the following: create an array of strings 0 to 99. The function should first check if input value is in the range and return null of something to indicate invalid input. If the value is in the valid range, return the array element by its index.

Simple as that.

—SA

Posted 14-Nov-11 9:04am

Sergey Alexandrovich Kryukov

Comments

MG2011 14-Nov-11 15:13pm

I haven' learnt how to make arrays yet... It sounds simple, but it's still theory to me.
Thanks anyway, though!

Sergey Alexandrovich Kryukov 14-Nov-11 19:19pm

You are welcome.

So, go to next step! It will work, believe me.
If you agree it makes sense, please formally accept the answer (green button) -- thanks.
--SA

MG2011 15-Nov-11 9:16am

OK I will try and make an array and I will post the outcome here, so that you can correct my mistakes. Oooh I'm anxious, this will be my first try! Anyway, it is definitely simplier with arrays, that the other way(That creates a loop for every ten numbers) so it's worth trying!

Sergey Alexandrovich Kryukov 15-Nov-11 11:23am

Great. Please don't post it as a solution; instead, use "Improve question" above and add a section to the body of your question (where you can format code as <pre lang="c">....</pre>). Post a comment here as well.
--SA

MG2011 19-Nov-11 9:34am

Well I tried sth..it's working but still I have to learn more thingsabout arrays to work with them easier and quicker..!
Here is the code:(I don't know how to make "eleve, twelve...nineteen" show up if they are in the array...I get some very funny results, like tentwo for twelve..lol!!!)
#include <stdio.h>
#include <stdlib.h>

main()
{
char ar1[11][12]={ "","ten","twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" };
char ar2[11][10]={ "", "one", "two", "three", "four", "five", "six","seven", "eight", "nine", "ten"};
int x, y, z;
printf("Give number: \n");
scanf("%d", &x);
z=x/10;
y=x%10;
if (z==0 && y==0)
{
printf("Zero.\n");
goto end;
}
if (y==1 && z==1)
{
printf("Eleven.\n");
goto end;
}
if (z==1 && y==2)
{
printf("Twelve.\n");
goto end;
}
if (z==1 && y==3)
{
printf("Thirteen.\n");
goto end;
}
if (z==1 && y==4)
{
printf("Fourteen.\n");
goto end;
}
if (z==1 && y==5)
{
printf("Fifteen.\n");
goto end;
}
if (z==1 && y==6)
{
printf("Sixteen.\n");
goto end;
}
if (z==1 && y==7)
{
printf("Seventeen.\n");
goto end;
}
if (z==1 && y==8)
{
printf("Eighteen.\n");
goto end;
}
if (z==1 && y==9)
{
printf("Nineteen.\n");
goto end;
}
printf("%s%s!\n", ar1[z],ar2[y]);
end:
system("pause");
}

Member 8393704 · Answer 4 · 2011-11-17T03:52:00

C#

void printnum(int num)
{
char *iOnesTens[] = { "zero", "one", "two", "three", "four", "five", "six",   "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };
char *iTens[] = { "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" };

if(num==0)
 printf(" %s",iOnesTens[num]);
else
{
 if( num / 20)
 {
  printf("\n %s",iTens[num/10]);
  num=num%10;
 }
 if(num!=0)
  printf(" %s",iOnesTens[num]);
}
}

Read a number, show me a word!

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